The $\mathbb{T}$-Cartier divisor $D_{\Psi}$ is given by $\{(X_{\tau},\chi^{-m_{\tau}})\}_{\tau\in\Sigma}$. If $m_{\sigma}=0$, the local equation of $D_{\Psi}$ in $X_{\sigma}$ is $\chi^{0}=1$. Therefore, the orbit $O(\sigma)$ does not meet the support of $D_{\Psi}$. Hence $V(\sigma)$ and $D_{\Psi}$ intersect properly.

To see that $\{m_{{\overline{\tau}}}\}_{{\overline{\tau}}\in\Sigma(\sigma)}$ is a set of defining vectors, we pick a point ${\overline{u}}\in{\overline{\tau}}$ and we choose $u\in\tau$ such that $\pi_{\sigma}(u)={\overline{u}}$. Then

$$\Psi(\sigma)({\overline{u}})=\Psi(u)=m_{\tau}(u)=m_{{\overline{\tau}}}({\overline{u}}),$$

which proves the claim. Now, using the characterization of $\Psi(\sigma)$ in terms of defining vectors, we have

$$\iota_{\sigma}^{\ast}D_{\Psi}=\{(X_{\tau}\cap V(\sigma),\chi^{-m_{\tau}}\mid_{X_{\tau}\cap V(\sigma)})\}_{{\overline{\tau}}}=\{(X_{{\overline{\tau}}},\chi^{-m_{{\overline{\tau}}}})\}_{{\overline{\tau}}}=D_{\Psi(\sigma)}.$$

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